Hugo's bike tire has a piece of gum stuck to it. The distance $G(t)$ (in $\text{cm}$ ) between the gum and the sidewalk as a function of time $t$ (in seconds) can be modeled by a sinusoidal expression of the form $a\cdot\sin(b\cdot t)+d$. At $t=0$, the gum is halfway between the ground and its maximum height, at $35\text{ cm}$. The gum reaches its maximum height of $70 \text{ cm}$ from the ground $\dfrac{\pi}{20}$ seconds later. Find $G(t)$. $\textit{t}$ should be in radians. $G(t) = $
Explanation: The strategy First, we should convert the given information about the real-world context into mathematical terms of the sinusoidal function and its graph. Then, we should use the given information to find the amplitude, midline, and period of the function's graph. Finally, we should find $a$, $b$, and $d$ in the expression $a\sin(b\cdot t)+d$ by considering the features we found. Converting the given information into mathematical terms At $t=0$, the gum is $35\text{ cm}$ above the ground. This means the graph of the function passes through $(0,35)$. We are given that this is halfway between the ground and its maximum height, which corresponds to the midline of the graph. $\dfrac{\pi}{20}$ seconds later (which means $t=\dfrac{\pi}{20}$ ) the distance is $70\text{ cm}$. This corresponds to the point $\left(\dfrac{\pi}{20},70\right)$. We are given that this is farthest point from the ground, which corresponds to a maximum point of the graph. In conclusion, the graph intersects its midline at $(0,35)$ and then has a maximum point at $\left(\dfrac{\pi}{20},70\right)$. Determining the amplitude, midline, and period The midline intersection is at $y={35}$, so this is the midline. The maximum point is $35$ units above the midline, so the amplitude is ${35}$. The maximum point is $\dfrac{\pi}{20}$ units to the right of the midline intersection, so the period is $4\cdot \dfrac{\pi}{20}={\dfrac{\pi}{5}}$. [Why did we multiply by 4?] Determining the parameters in $a\sin(b\cdot t)+d$ Since the midline intersection at $t=0$ is followed by a maximum point, we know that $a>0$. [How do we know that?] The amplitude is ${35}$, so $|a|={35}$. Since $a>0$, we can conclude that $a=35$. The midline is $y={35}$, so $d=35$. The period is ${\dfrac{\pi}{5}}$, so $b=\dfrac{2\pi}{\left({\dfrac{\pi}{5}}\right)}=10$. The answer $G(t)=35\sin\left(10t\right)+35$